1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bead attached to a spring and moving along a horizontal wire

  1. Dec 1, 2012 #1
    I'm self-studying an introductory book on mathematical methods and models and came across the following problem:

    1. A bead of mass m is threaded onto a frictionless horizontal wire. The bead is attached to a model spring of stiffness k and natural length l0, whose other end is fixed to a point A at a vertical distance h from the wire (where h > l0). The position x of the bead is measured from the point on the wire closest to A. Find the potential energy function U(x).


    2. Relevant equations

    I'm rather puzzled by the solution given in the book, which claims that since the length of the spring is (h2+x2)1/2 and its extension is (h2+x2)1/2 - l0, then U(x) = (1/2)k((h2+x2)1/2 - l0)2.


    3. The attempt at a solution

    I think that's incorrect because only the x-component of the force exerted by the spring on the bead is relevant to the calculation of U(x). The x-component is -k(l-l0)cos[itex]\theta[/itex], where l = (h2+x2)1/2 and cos[itex]\theta[/itex] = x/(h2+x2)1/2. Hence, U(x) is -[itex]\int[/itex](-kx + kl0x/(h2+x2)1/2)dx. Is this correct?
     
    Last edited: Dec 1, 2012
  2. jcsd
  3. Dec 1, 2012 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Humm. Interesting question. I'm not sure I can answer it. At first thought you are correct, but then ...

    With the spring stretched the energy stored in the spring will be that calculated in the book. What happens to the energy when the spring is released? Where does it go if you apply conservation of energy? The wire is frictionless so as far as I can see all the energy ends up in the bead (eg not some fraction of that due to the Cosθ issue you raise).

    So I conclude the book is correct but I'm willing to be convinced you are right and the book is wrong.
     
  4. Dec 1, 2012 #3

    TSny

    User Avatar
    Homework Helper
    Gold Member

    See if you can show that both expressions are correct, although they might differ by an additive constant.
     
  5. Dec 1, 2012 #4
    The integral should evaluate to

    kx2/2 - kl0(h2 + x2)1/2 + C.

    Hence, the difference between my solution and the book's is:

    kx2/2 - kl0(h2 + x2)1/2 + C - (k(h2 + x2)1/2l0 + (1/2)kh2 + (1/2)kx2 - (1/2)kl02)
    = C - (1/2)kh2 + (1/2)kl02,

    so the two solutions do appear to differ by a constant...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook